A computer uses a memory unit of 512 K words of 32 bits each. A binary instruction code is stored in one word of the memory. The instruction has four parts: an addressing mode field to specify one of the two-addresssing mode (direct and indirect), an operation code, a register code part to specify one of the 256 registers and an address part. How many bits are there in addressing mode part, opcode part, register code part and the address part?

1. A1, 3, 9, 19
2. B1, 4, 9, 18
3. C1, 4, 8, 19
4. D1, 3, 8, 20

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Correct answer: C. 1, 4, 8, 19

Explanation

512K words needs 19 address bits; 256 registers need 8 bits; two addressing modes need 1 bit; the opcode gets the rest: 32 − 19 − 8 − 1 = 4.